Question

If $1,w,w^2$ are cube roots of unity then $(a+b{)}^3+(aw+b{w}^2{)}^3+(a{w}^2+bw{)}^3=$

• A. $a^3+b^3$
• B. $3(a^3+b^3)$
• C. $a^2-b^3$
• D. $a^3+b^3+3ab$

Answer 1

Answer: (B)

$3(a^3+b^3)$

Taking
$x=a+b$
$y=a\omega +b{\omega }^2$
$z=a{\omega }^2+b\omega$
We have
$x+y+z=a(1+\omega +{\omega }^2)+b(1+\omega +{\omega }^2)$$=0$ as $1+\omega +{\omega }^2=0$
So,
$x^3+y^3+z^3$

$=3xyz$
$=3(a+b)(a\omega +b{\omega }^2)(a{\omega }^2+b\omega )$
$=3(a+b)(a^2-ab+b^2)$
$=3(a^3+b^3)$